A wave y = a cos (kx + omega t) superimposes | vedclass

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(C) stationary wave is formed by the superposition of two waves of the same frequency and amplitude traveling in opposite directions.Given the first w

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$ - a \cos (kx - \omega t)$

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(C) stationary wave is formed by the superposition of two waves of the same frequency and amplitude traveling in opposite directions.Given the first wave is $y_1 = a \cos (kx + \omega t)$.For a stationary wave to have a node at $x = 0$,the resultant displacement $y = y_1 + y_2$ must be zero at $x = 0$ for all $t$.So,$y_1(0, t) + y_2(0, t) = 0$.$a \cos (\omega t) + y_2(0, t) = 0$,which implies $y_2(0, t) = -a \cos (\omega t)$.Since the waves must travel in opposite directions,if the first wave is traveling in the negative $x$-direction (as indicated by $kx + \omega t$),the second wave must travel in the positive $x$-direction,taking the form $y_2 = a' \cos (kx - \omega t + \phi)$.At $x = 0$,$y_2(0, t) = a' \cos (- \omega t + \phi) = a' \cos (\omega t - \phi)$.Comparing this with $-a \cos (\omega t)$,we get $a' = a$ and $\phi = \pi$.Thus,$y_2 = a \cos (kx - \omega t + \pi) = -a \cos (kx - \omega t)$.

$A$ wave $y = a \cos (kx + \omega t)$ superimposes on another wave to form a stationary wave having a node at $x = 0$. What is the equation of the other wave?

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